A 5.00 g sample of water vapor, initially at 155degreeC is cooled at atmospheric

Question

A 5.00 g sample of water vapor, initially at 155degreeC is cooled at atmospheric pressure, producing ice at -55degreeC. Calculate the amount of heat energy lost by the water sample in this process, in kJ. Use the following data: specific heat capacity of ice is 2.09 J/g-K; specific heat capacity of liquid water is 4.18 J/g-K; specific heat capacity of water vapor is 1.84 J/g-K; heat of fusion of ice is 336 J/g; heat of vaporization of water is 2260 J/g. Draw heating curve diagram with labeled axes and show work

Explanation / Answer

heat changed= water is formed from vapour+water cooled to 0^oC + water changed to ice+icecooled to -55^0C=mx Lv+ mCwdeltaT + mLi + mCideltaT

5x4.18 x- 100- 5x336+ 5x 2.09 x-55=-2090-1680 -574.75=-4344.75

Energy of vapour=5x2260+ 5x 1.84 x55 =11300+506=11806J

heat lost=11806-4344.75=7461.25= 7.461KJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.